Problems

Among the first \(20\) Fibonacci numbers: \(F_0 = 0,F_1 = 1,F_2 = 1, F_3 = 2, F_4 = 3,..., F_{20} = 6765\) find all numbers whose digit-sum is equal to their index. For example, \(F_1=1\) fits the description, but \(F_{20} = 6765\) does not, since \(6+7+6+5 \neq 20\).

Among the first \(20\) Fibonacci numbers: \(F_0 = 0,F_1 = 1,F_2 = 1, F_3 = 2, F_4 = 3,..., F_{20} = 6765\) check whether the numbers with prime index are prime. The index is another name for a number’s place in the sequence.

Consider Pascal’s triangle: it starts with \(1\), then each entry in the triangle is the sum of the two numbers above it. Prove that the diagonals of Pascal’s triangle add up to Fibonacci numbers.

Prove for any \(m,n\) that \(F_{m+n} = F_{m-1}F_n + F_mF_{n+1}\).
Corollary: if \(k\mid n\), then \(F_k\mid F_n\). This can be proven by induction if we write \(n=sk\) for a natural \(s\), then \[F_{k+(s-1)k} = F_{k-1}F_{(s-1)k} + F_kF_{(s-1)k+1}.\]

Denote by \(\gcd(m,n)\) the greatest common divisor of numbers \(m,n\), namely the largest possible \(d\) which divides both \(n\) and \(m\). Prove for any \(m,n\) that \[\gcd(F_n,F_m) = F_{\gcd(m,n)}.\]

Have you wondered if \(F_{-5}\) is possible? Here is how we can extend the Fibonacci sequence to the negative indices. The relation \(F_{n+1} = F_n + F_{n-1}\) can be rewritten as \(F_{n-1} = F_{n+1} - F_n\). We can simply define the Fibonacci sequence with negative indices with this formula. For example, \(F_{-1} = F_1 - F_0 = 1 - 0 = 1\).

Write out \(F_{-1}, F_{-2},\dots,F_{-10}\). What do you notice about the Fibonacci sequence with negative indices?